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Relating integral of powers of Sin b/w 0 and pi/2 to factorial form

  1. Aug 13, 2014 #1
    Our integral
    [tex]\int\limits_0^{\pi/2} \sin^{2a+1}(x)\,dx[/tex]

    Has a Factorial Form:
    [tex] {(2^a a!)}^2 \over (2a+1)![/tex]

    What is the process behind going from that integral to that factorial form?

    My approach which is not very insightful:
    I used mathematica to calculate the integral to return:
    [tex]\pmb{\frac{\sqrt{\pi } \text{Gamma}[1+a]}{2 \text{Gamma}\left[\frac{3}{2}+a\right]}}[/tex]
    I know Gamma[1+a] = a! and Gamma[3/2+a] has a factorial form also but doesn't help me to reduce to that form.

    Griffiths just says that integral equals (2*4*...2a)/(1*3*5...*[2a+1]) to get from this to that factorial form is easy but I got lost in his integration.
     
    Last edited: Aug 13, 2014
  2. jcsd
  3. Aug 13, 2014 #2

    micromass

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    What do you get after integrating by parts twice?
     
  4. Aug 13, 2014 #3
    [tex]\pi/2-(2a+1)\int\limits_0^{\pi/2} \sin^{2a}(x)cos(x)x[/tex]

    Sorry I still don't see how to finish the connection.

    Edit: Didn't see you said twice IBP ill go back and retry this
     
  5. Aug 13, 2014 #4
    taking u = sin^2a(x) and v'=xcos(x) , I get:
    [tex] -a\pi+(2a+1)(2a)\int\limits_0^{\pi/2} Sin^{2a}(x)x +Sin^{2a-1}(x)cos(x)dx[/tex]

    Sadly my math isn't great and cant seem to figure out how this would lead to the factorial form.

    So I tried to integrate by parts again and try to simplify and it just got really messy.
    I tried to take u = sin^2a(x)cos(x) and v'=x and that got messy also compared to the one above so I didn't pursue it.
     
    Last edited: Aug 13, 2014
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